Homomorphism of groups

Question

Let $\phi : G_1 \rightarrow G_2$ be a homomorphism of groups.

Now I have to prove that for any $g \in G_1$ we have $\phi (g^{-1}) = [\phi (g)]^{-1}$.

So how should I begin?

Answer 1

Hint: what's the defining property of an inverse? Can you use the properties of homomorphisms to show that $\phi(g^{-1})$ satisfies the relevant property?

Answer 2

For any $g\in G$ we have $gg^{-1}=1=g^{-1}g$

For any homomorphism $\phi:G\rightarrow G'$ we have

$$\phi(g)\phi(g^{-1})=\phi(gg^{-1})=\phi(1)=1=\phi(1)=\phi(g^{-1}g)=\phi(g^{-1})\phi(g)$$